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stiks02 [169]
2 years ago
12

Which is the least likely cause of an engine to hunt and surge at top no-load speeds? A lean air/fuel mixture An incorrect spark

plug heat range. A blocked carburetor bowl vent All of the above.
Physics
2 answers:
77julia77 [94]2 years ago
8 0
I would say that the answer to this is the last option: ALL OF THE ABOVE. A lean fuel or air mixture, incorrect spark plug heat range, and a blocked carburetor bowl vent would be among the reasons of an engine to surge and hunt at the top no-load speeds. Hope this answer helps.
Zolol [24]2 years ago
4 0

Answer:

A blocked  carburetor bowl vent

Explanation:

Engine hunt is defined as the continuous  variation in rpm of the engine. It can occur due to various conditions and abnormalities in the inside the fuel injector pump.

In ideal situation, amount of fuel given to the injector is fixed  but in some cases such abnormality arises hence rpm get changed.

Hence of the options available, a blocked carburetor bowl is the least likely cause for engine hunt.

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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
babunello [35]

Answer:

Separation increases at all times that rock X falls because it falls with a greater speed

Explanation:

For both rocks, let initial velocity ∪=0

To find the displacement at any given time interval of Δt then

S= ∪Δt +0.5gΔt²

Since rock X is first released followed by Y, then X has a greater speed than Y therefore the distance covered by X is longer. This is because despite 0.5gΔt² being same for both rocks at any time Δt but rock X having already attained some velocity, its ∪Δt  is more hence the separation S increases. Conclusively, S increases at all times that rock X falls since rock X falls with a greater velocity than rock Y

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2 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

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